It is found that for every $10^{\circ}C$ rise in temperature,the rate of the reaction doubles itself. This is mainly because

In a reaction, for every $10^{\circ} C$ rise of temperature, the rate is doubled. If the temperature is increased from $10^{\circ} C$ to $100^{\circ} C$, the rate of the reaction will become $:-$ (in $\times$)

What is the activation energy for a reaction if its rate doubles when the temperature is raised from $20 \,^{\circ}C$ to $35 \,^{\circ}C$ in $kJ \,mol^{-1}$? $(R = 8.314 \,J \,mol^{-1} \,K^{-1})$

The decomposition of ethane,$\frac{d[C_2H_6]}{dt} = k[C_2H_6]$,proceeds through a complex mechanism,which includes $5$ steps. The overall rate constant $(k)$ is expressed as $k = \frac{k_1 k_2 k_3}{k_2 k_5}$,where $k_1, k_2, k_3, k_4, k_5$ are the rate constants of the $5$ steps. If the activation energies of the steps are $E_1 = 1E, E_2 = 2E, E_3 = 3E, E_4 = 4E, E_5 = 5E$,where $E = 20 \ kJ/mol$,find the overall activation energy of the decomposition.

The rate constants $k_1$ and $k_2$ for two different reactions are $10^{16} \times e^{-2000/T}$ and $10^{15} \times e^{-1000/T}$ respectively. The temperature at which $k_1 = k_2$ is

$A$ catalyst lowers the activation energy for a certain reaction from $83.314 \, kJ \, mol^{-1}$ to $75 \, kJ \, mol^{-1}$ at $500 \, K$. What will be the rate of reaction as compared to the uncatalysed reaction? Assume other things being equal.